85

6 Thermal properties

Var . Dim. Description

∆r/a −− fractional expansion

a L bond length at T = 0

ǫc ML2T−2 bond energy

kT ML2T−2 thermal energy

Table 6.1. Variables that determine

thermal expansion. The goal, ∆r/a, is

in red.

How fast does a turkey cook? How quickly does the moon cool? Why

are wooden spoons useful for cooking? Why does water boil at 373K?

Such questions depend on the thermal properties of materials. To

warm up, try thermal expansion.

6.1 Thermal expansion

Most materials expand when they get hotter and contract when they

get cooler. If bridge designers forget this fact, and design a bridge so

that its joints exactly mate in the summer, then it will fall down in

the winter. We would like to understand why materials expand, and

by how much.

So imagine a piece of material, heated from 0K to a temperature

T . How much does it lengthen? Before trying dimensional analysis,

you can sharpen the question. Materials lengthen because their atoms

move apart. So a long rod probably expands more than a short rod

does. A physically reasonable quantity to analyze is

change in length

length, (6.1)

which is the fractional length change. This quantity is intensive, usu-

ally a good quality, and dimensionless, always a good quality. So the

revised question is: In heating the material from 0K to temperature

T , by what fraction does it lengthen? Time for the list of variables.

With a as the usual bond length, let ∆r be the change in length. Then

∆r/a is the fractional change. As the goal, it goes on the list. The

temperature is on the list. As a rule of thumb, include temperature as

the energy kT , where k is Boltzmann’s constant. If you include it as

T , you add a fourth dimension, temperature. To deal with a fourth di-

mension, you then need to include a constant to convert temperature

to energy; that constant is k. Including T and k separately adds two

variables and one dimension, which increases the number of dimen-

sionless groups by one. Equivalently – and more simply – include kT ,

which adds one variable and no new dimensions. This simpler alter-

native also adds also one new dimensionless group. So kT goes on the

list. Strong bonds imply a stiff, hard-to-stretch material, so the bond

strength – or cohesive energy ǫc – might affect thermal expansion as

well. Perhaps the bond length a matters as well. The list is shown in

Table 6.1.

2006-01-22 12:49:46 [rev 251865795177]

6. Thermal properties 86

Cu

Hg

Si Fe

AlH2O

Pb

Au

Fe

αLvap

R= 1

αLvap

R= 0.1

2 5 10 20 50 100

0.1

0.2

0.5

1

2

5

10

20

50

α (10−6 K−1)

Lvap

R(104 K)

Figure 6.1. Cohesive energy vs ther-

mal expansion coefficient. The cohe-

sive energy is represented by Lvap/R,

where Lvap is the molar heat of vapor-

ization. The solid red line is the curve

α(Lvap/R) = 1 and the dotted red line

is α(Lvap/R) = 0.1. Most substances lie

between these lines.

Although this list contains three dimensions (L, M, and T), it con-

tains only two independent dimensions. You can form the dimensions

of every variable using only L and ML2T−2. Four variables and two

independent dimensions produce two dimensionless groups. The first

one is free because ∆r/a is already dimensionless. So

Π1 =∆r

a. (6.2)

The other group is almost as easy. Both ǫc and kT are energies, so

Π2 =kT

ǫc(6.3)

is dimensionless. Two groups complete the set. Only one variable, a,

has dimensions of length, so a does not make it into a dimensionless

group. To solve for ∆r/a, use the form Π1 = f(Π2):

∆r

a= f

(kT

ǫc

)

. (6.4)

A reasonable guess is that expansion is proportional to temperature,

so f(x) would be linear. Then

∆r

a∼ kT

ǫc=

k

ǫcT. (6.5)

where the factor in red is the thermal expansion coefficient α.

Then

α ∼ k

ǫc. (6.6)

The coefficient varies with temperature. Rather than changing the

temperature from 0 to T , over which range the coefficient might vary,

imagine a small temperature change ∆T around the current temper-

ature. The more exact definition of α uses ∆T instead of T :

α ≡ fractional length change

∆T. (6.7)

In an approximate analysis, you can often ignore the distinction be-

tween T and ∆T .

2006-01-22 12:49:46 [rev 251865795177]

6. Thermal properties 87

α Lvap/R

Substance (10−6 K−1) (104 K)

Au 14.2 3.90

Al 23.1 0.350

Cu 16.5 3.61

Si 2.6 4.32

Fe 11.8 4.20

Pb 28.9 2.14

Hg 60.7 0.71

Steel (carbon) 10.7

C

diamond 1.0graphite 7.1

Wood

along gr. 3–6

against gr 35–60

Water 70 0.500

Granite 4–7

Brick 3–10

Cement 7–14

Glasses

Pyrex 3.2

crown 7–8flint 8–9

Vycor 0.75

Table 6.2. Thermal-expansion coeffi-

cients at room temperature. The third

column gives, where available, ǫc in tem-

perature units:

ǫc

k×

NA

NA

=Lvap

R. (6.8)

For the liquids (water and mercury),

the tables quote the volume-expansion

coefficient; the linear-expansion coeffi-

cient above was obtained by dividing by

3. Source: Kaye and Laby [30, §2.3.5,

3.10.1, 3.11.2], Elements [13].

Table 6.2 lists thermal-expansion coefficients for various materi-

als, and Figure 6.1 shows how the data compare to the dimensional-

analysis prediction (6.6). If the missing constant in (6.6) is β, then

all substances would lie on the line α(Lvap/R) = β. Most substances

shown on the scatterplot have values between 0.1 (dotted red line)

and 1 (solid red line). So dimensional analysis produces a reasonable

result. Even using a fixed value of ǫc, say ǫc = 10 eV, in (6.6) produces

a reasonable, easy-to-remember value:

α ∼

k︷ ︸︸ ︷

10−4 eV K−1

10 eV∼ 10−5 K−1. (6.9)

Beyond dimensional analysis, you might wonder about the mech-

anism of thermal expansion. The mechanism also shows where the

missing constant in (6.6) comes from. Look at the potential energy

between two atoms or molecules (as in water). When T = 0 the atoms

have zero kinetic energy, the bond does not vibrate, and the atoms are

separated by a distance a (the minimum-energy distance). Quantum

mechanics prevents the kinetic energy from being zero, even at T = 0,

but that complication does not change the reasoning, so ignore the

so-called zero-point energy. Once you are happy with the rest of the

argument, you can improve the argument by adding it back.

As T increases so does the kinetic energy of the atoms. The bond

vibrates, and the potential-energy curve says how much. Think of the

bond as a vibrating spring. It expands until all kinetic energy converts

to potential energy. Without kinetic energy, it stops expanding; that

point is the maximum extension. The restoring force then pulls the

bond back toward its equilibrium length a, but it overshoots (like any

spring) and keeps shrinking until, again, all kinetic energy converts

to potential energy. That point is the minimum extension. A reason-

able guess is that the bond’s average extension is midway between

the minimum and maximum extensions. Figure 6.2 shows these mid-

way points for a true spring potential. The points do not vary with

temperature, so there is no thermal expansion! Thermal expansion

requires an asymmetric potential, such as the Lennard–Jones poten-

tial (Figure 6.3) or the potential (5.14) for hydrogen including the

confinement energy.

Figure 6.3 shows the midpoints shifting with increasing tempera-

ture. To find an expression for their position, solve for the minimum

and maximum distances as a function of kinetic energy. The kinetic

energy is roughly kT , so kT = U(r±) − U(a), where r+ is the max-

imum extension and r− is the minimum extension. This equation is

easier to analyze in dimensionless form. It has lengths and energies,

and they have a natural scale: ǫc for energy and a for length. So use

2006-01-22 12:49:46 [rev 251865795177]

6. Thermal properties 88

r

U

a

Figure 6.2. Symmetric potential. As the

temperature increases (horizontal lines),

the bond oscillates between two extreme

lengths. In this symmetric spring poten-

tial, the point (in red) midway between

these extremes does not vary with tem-perature.

r

U

a

Lennard–Jones

Figure 6.3. An asymmetric potential

(the Lennard–Jones potential). Here the

midpoint (in red) varies with tempera-

ture.

a and ǫc to define dimensionless variables:

T ≡ kT

ǫc,

U ≡ U

ǫc+ 1,

r ≡ r

a− 1.

(6.10)

The definition of U includes a +1 so that the minimum U is 0 rather

than -1. Similarly, the definition of r includes a −1 so that U has

a minimum at r = 0 instead of at r = 1. This choice – placing

a special point (the minimum) at a special location (the origin) –

makes subsequent calculations slightly simpler and shorter. With this

notation, the problem becomes one of solving

T = U(r). (6.11)

Around its minimum, any function looks like a parabola. So

U(r) = βr2 + · · · , (6.12)

where β is a constant. Without the r3 term in the dots, the U is

symmetric and the substance does not expand thermally (Figure 6.2).

A Taylor series produces the r3 term:

U(r) =1

2U ′′(0)r2 +

1

6U ′′′(0)r3 + . . . . (6.13)

The constant term vanishes because U is defined so that U(0) = 0.

The linear term U ′(0)r vanishes because the expansion is about the

minimum, where the first derivative is zero.

In this dimensionless world, U ′′(0) and U ′′′(0) are dimensionless

constants. Follow the usual practice and pretend that they are 1. Bet-

ter: Pretend that U ′′(0) = 2 and U ′′′(0) = −6, to make the formula

as simple as possible. Choose U ′′′(0) to be negative so that U be-

haves correctly: Compared to a pure spring (U = r2), the true U has

stronger repulsion at short distances and weaker attraction at long

distances. With these choices,

U(r) ∼ r2 − r3. (6.14)

This function is sketched and analyzed in Figure 6.4. The result is that

the equilibrium point of the bond shifts to r = T /2. The definition of

r is r = r/a − 1, so r = T /2 becomes ∆r/a = T /2. Since T = kT/ǫc,

the result in regular dimensions is:

∆r

a∼ 1

2

k

ǫcT. (6.15)

2006-01-22 12:49:46 [rev 251865795177]

6. Thermal properties 89

r

U

r3

?

r2

r2− r3

T

Figure 6.4. Rescaled (dimensionless)

potential energy. The red curve includes

the −r3 term. The black curve is the

symmetric potential energy. The thermal

expansion due to a scaled temperatureT is caused by the difference between

these curves, and it is the average of the

two little horizontal lines. For the right-

hand line, the height of the triangle is r3

because it is the difference between the

symmetric r2 curve and the asymmetric

r2 − r3 curve. The slope of the symmet-

ric curve is 2r, so the horizontal piece

has length r3/2r = r2/2. Since T ≈ r2

(except for the r3 term, which is small),

the length is T /2. By symmetry, the left-hand shift is also T /2, so the midpoint

shifts by T /2 going from the black curve

to the red curve.

ǫc Tvap (K)

Substance (eV) actual pred .

Water 0.50 373 580

NH3 0.31 240 348

HCl 0.21 188 244

O2 0.071 90 82Au 3.35 3130 3884

Xe 0.13 165 150

He 0.00086 4.2 1

Hg 0.61 630 707

N2 0.058 77 67

Table 6.3. Cohesive energy, ǫc, per

atom or molecule; actual and predicted

boiling temperatures (at 1 atm). The co-

hesive energy is Lvap/NA, where Lvap

(the latent heat of vaporization) is from

experimental data. The predicted boilingpoint comes is ǫc/10k, the prediction of

(6.20). Source: CRC [41, 6-106ff]

The thermal expansion coefficient is everything on the right except

for the T (in red), so

α ∼ 1

2

k

ǫc. (6.16)

To play fair we should drop the 1/2, as with the other dimensionless

constants. But including the 1/2 makes the thermal-expansion pre-

diction too accurate for us to resist temptation. In Figure 6.1, the line

α(Lvap/R) = 1/2 would pass very near most of the points.

The analysis in Figure 6.4 assumes that r is small. Is this assump-

tion justified? Although the analysis using a spring potential does not

explain thermal expansion, it does give us the estimate that r ∼√

T

(in dimensionless units). A typical kT is 0.025 eV; a typical covalent

bond energy is ǫc ∼ 2.5 eV. In dimensionless units, T ∼ 0.01 and

r ∼ 0.1. For most order-of-magnitude calculations, 0.1 is a small di-

mensionless number. Here its smallness justifies the approximations

made in analyzing Figure 6.4.

6.2 Boiling

To vaporize or boil a liquid requires energy to move molecules from the

liquid into the gas phase. How much energy is required? Per molecule,

the energy required is roughly the cohesive energy ǫc. The molar heat

of vaporization (or enthalpy of vaporization) is the energy required to

boil one mole of a substance, and we can estimate it from the cohesive

energy:

Lvap ∼ NAǫc ∼ 100 kJ mol−1 ×( ǫc

1 eV

)

. (6.17)

For water, ǫc ∼ 0.5 eV so Lvap ∼ 50 kJ mol−1. This estimate of the

enthalpy leaves out a small contribution: the energy to make room

in the vapor for the evaporating molecules. This energy is PV in the

expression H = E + PV for the enthalpy. The heat of vaporization is

the change in H:

Lvap ≡ ∆H = ∆E + ∆(PV ). (6.18)

Once you look at data for boiling temperature, you find that ∆(PV )

is small compared to ∆E. But dot the i’s at the end!

Predicting Lvap from ǫc is straightforward. Instead use Lvap, for

which accurate experimental data are available, to determine ǫc (the

second column of Table 6.3). Now knowing ǫc you can predict the

boiling temperature. As a substance heats up, more molecules leave

the liquid and become vapor; the pressure of the vapor increases. The

boiling temperature is the temperature at which the vapor pressure

reaches atmospheric pressure (at sea level, roughly 105 Pa or 1 atm).

Atmospheric pressure is arbitrary; it is lower on Mount Everest, for

example, and far lower on Mars. It is unrelated to the properties of the

substance being boiled – although it helps to determine the boiling

2006-01-22 12:49:46 [rev 251865795177]

6. Thermal properties 90

temperature. A more general question is how boiling temperature

depends on atmospheric pressure. You can generalize the upcoming

methods to answer this question.

For a first guess for the boiling temperature, convert the cohesive

energy into a temperature using Boltzmann’s constant:

Tvap = Πǫc

k, (6.19)

where Π is a dimensionless constant. The conversion factor k is 1 eV ≃104 K (accurate to 20 percent). Table 6.3 shows how inaccurate this

guess is. For water, the predicted boiling temperature would be 5000K

rather than 373K: Tea kettles would be difficult to boil. For gold, the

predicted temperature would be 30, 000K instead of ∼ 3000K. So by

fiat insert a factor of 10 to improve the prediction to:

Tvap ∼ ǫc

10k, (6.20)

If Tvap, ǫc, and k are the only relevant variables for the boiling tem-

perature, then there is a constant Π such that Tvap = Πǫc/k. A few

minutes with the data in Table 6.3 will convince you that, although

the factor is nearly 0.1 as in (6.20), it is not a constant. However, even

the approximate result helps check that PV can be neglected in the

heat of vaporization (6.18). For one molecule in a gas at the boiling

temperature, PV = kTvap. From (6.20), kTvap ∼ ǫc/10 so PV is also

ǫc/10. Thus PV is much smaller than ǫc, which is the consistency

check that we promised.

6.3 Heat flow

The previous sections discussed what happens when you heat or cool

a substance. Heating or cooling requires sending heat from one sub-

stance to another. But how does heat travel? And how fast? These

questions lead to the subject of heat flow, upon which many interest-

ing phenomena depend: Why can people (who are not too confident

of success) walk on hot coals without getting burned? How long does

the 3-inch layer of frost in the freezer take to melt? Heat flow is the

topic of the rest of the chapter.

Charge flow (current) and heat flow share a key property: They

flow in response to a drive. Charge flows from high potential to low po-

tential. Heat flows from high temperature to low temperature. Charge

flow is measured by current density, also known as charge flux:

charge flux =charge transported

area × time. (6.21)

That is the definition of charge flux. It is produced by an electric field:

charge flux︸ ︷︷ ︸

J

= conductivity︸ ︷︷ ︸

σ

× drive︸ ︷︷ ︸

E

. (6.22)

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6. Thermal properties 91

T1 T2

∆x

Figure 6.5. Slab of gas. The slab hascross-sectional area A (not shown). As

molecules from the left side diffuse into

the right side, and vice versa, the tem-

peratures equalize. The heat flow is from

hot to cold, and can be estimated from

knowing the heat content in each half

and the time it takes to diffuse.

The electric field E is produced by the potential difference and is

driving the charges. Imagine a slab of material, perhaps a wire or a

block of seawater (Section 4.6), of length ∆x. Then the electric field

is

E =V1 − V2

∆x, (6.23)

where V1 and V2 are the potentials at the ends. Imagine heat flowing

in the same slab. Heat flow is measured by heat flux,

heat flux =energy transported

area × time. (6.24)

The analogy to heat flow suggests a form for heat flux similar to the

charge flux (6.22):

heat flux︸ ︷︷ ︸

F

= thermal conductivity︸ ︷︷ ︸

K

× drive︸ ︷︷ ︸

?

. (6.25)

The following symbols are conventional: F for flux and K for thermal

conductivity. Again by analogy to charge flow, the drive for heat flow

should be similar to the drive (6.23) for charge flow, so try

drive =T1 − T2

∆x. (6.26)

Then the heat flux (6.25) becomes

F = KT1 − T2

∆x. (6.27)

This argument by analogy does not explain the mechanism, but it

structures the analysis of the thermal conductivity.

The mechanism is easiest to understand in gases, so study them

first. Studying gases is also enjoyable: We are surrounded by them

(air), so their properties explain many everyday thermal phenomena

(feeling cold standing outside in a thin shirt, for example).

Fluxes, by construction, divide out the cross-sectional area. How-

ever, you can understand phenomena more easily with a mental pic-

ture, and it is difficult to imagine a substance without cross-sectional

area. So imagine a slab of gas with cross-sectional area A, knowing

(hoping, expecting) that A cancels when computing any intensive

quantity such as thermal conductivity. The length ∆x is already re-

quired, in order to compute the drive (T1−T2)/∆x, where the slab has

temperature T1 in the left half and T2 in the right half (Figure 6.5).

To find K, use this model to estimate the flux F , then match the

result against the general form (6.27). To compute the flux using

(6.24) requires knowing the: (1) energy transported, (2) time taken,

and (3) cross-sectional area. The area A is specified in the thought

experiment. The other two require further analysis.

2006-01-22 12:49:46 [rev 251865795177]

6. Thermal properties 92

Energy is transported because, after waiting a while, the temper-

atures equalize. So energy has flowed from left to right. This energy

is, except for dimensionless constants (ignore them!),

∆E ∼ energy in left side − energy in right side. (6.28)

Each portion can be estimated by estimating the energy per molecule.

This so-called internal energy is a function of T , call it u(T ), where

u is the traditional label for internal energy, and it is in lower case

to indicate that it is the energy per molecule. Each side contains

N ∼ nA∆x molecules, where n is their number density. So

∆E ∼ (u(T1) − u(T2))nA∆x. (6.29)

As long as u has constant slope, or T1 and T2 are close, this becomes

∆E ∼ ∂u

∂T(T1 − T2)nA∆x. (6.30)

The factor T1 − T2 has appeared! The factor ∂u/∂T appears so often

that it has a name, the specific heat:

cp ≡ ∂u

∂T. (6.31)

The ‘p’ in the subscript is for pressure, meaning that this quantity

is measured by heating a gas while keeping the pressure fixed (the

alternative is to keep the volume fixed, which gives cv). Then

∆E ∼ cpnA∆x(T1 − T2). (6.32)

Now for the time required to transport this much energy. The

energy is transported by the molecules moving from one side to the

other. How long before the two sides mix? Except for a dimensionless

constant, this time is the same as the time for a molecule to wander

from one side to the other. We say ‘wander’, because the molecules

do a random walk. In a random walk of N steps, the molecules travel

a typical distance ∝√

N . This result is fundamentally different from

a regular walk. In a regular walk, x ∝ t, where t is the time taken to

make N steps. The interesting quantity is not x or t itself, since they

can grow without limit, but their ratio x/t, also known as the speed.

In a random walk, where 〈x2〉 ∝ t, the interesting quantity is 〈x2〉/t,which is the analogue of speed for a random walk. It is the diffusion

constant and has dimensions of L2T−1. For more on random walks,

see the classic book Random Walks in Biology [3], full of fascinating

examples.

Since the gas molecule walks randomly, the time it requires to

cross from one side to the other, a distance ∆x, is

t ∝ (∆x)2. (6.33)

2006-01-22 12:49:46 [rev 251865795177]

6. Thermal properties 93

The constant that makes the dimensions correct is κ, the diffusion

constant:

t ∼ (∆x)2

κ. (6.34)

With this time and the energy transported (6.32), the heat flux (6.24)

becomes

F =∆E

At∼

∆E︷ ︸︸ ︷

cpnA∆x(T1 − T2)

A (∆x)2/κ︸ ︷︷ ︸

t

. (6.35)

As expected, the area cancels and

F = ncpκ︸ ︷︷ ︸

K

× T1 − T2

∆x︸ ︷︷ ︸

drive

. (6.36)

Matching this form to the form for the flux (6.27) gives

K = ncpκ. (6.37)

Let’s check upstairs and downstairs. As the gas becomes concentrated,

it contains more energy so it can transport energy more rapidly: n

should be on top. If each molecule’s capacity to hold energy increases,

then the same diffusive wandering transports more energy: cp should

be on top. As the gas molecules diffuse more easily, they transport

energy more rapidly: κ should be on top. The thermal conductivity

(6.37) passes these tests. The problem of heat flow has simplified to

problem of finding K. The problem of finding K is the problem of

finding the specific heat cp and the diffusivity κ. This thought exper-

iment used a gas, and the same ideas will apply for liquids and solids.

We continue with the thermal conductivity of gases by estimating cp

and κ for gases. After that we turn to liquids and solids.

6.4 Diffusivity of gases

The diffusion constant, or diffusivity, measures how rapidly the gas

molecules wander. Molecules travel until they hit another one, then

they bounce in a random direction: These ingredients make a random

walk. All we have to know is how far they wander before colliding,

and how often they collide. These natural units of space and unit

time characterize the random walk; the nature of the particle doing

the walking doesn’t matter beyond those two facts. Try dimensional

analysis. The relevant variables are κ, the goal; the collision time τ ;

and the mean free path λ. Three variables containing two independent

dimensions (L and T) make one dimensionless group: κτ/λ2. So

κ ∼ λ2

τ. (6.38)

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6. Thermal properties 94

In terms of the molecular velocity v = λ/τ ,

κ ∼ vλ. (6.39)

The missing constant depends a more exact calculation of the behav-

ior of a random walk, and also on the definition of κ. Several defi-

nitions are possible, each with its own dimensionless constant buried

in the definition. So these variations, not only the approximations in

the analysis, affect the constant. The convention for κ is to make the

diffusion equation,

∇2T = −κ

∂T

∂t, (6.40)

look elegant by not having a dimensionless constant (or rather by

making the constant unity). Using this definition of κ, a proper cal-

culation gives:

κ =1

3vλ, (6.41)

where the magic constant is in red. Here v is the thermal velocity

obtained from

thermal energy =1

2mv2

t , (6.42)

where m is the molecular mass. The thermal energy is (3/2)kT , so

vt =

√

3kT

m. (6.43)

For air, which is mostly diatomic nitrogen, m = 28mp. The speed

of sound (within 15 percent) is

cs ≈√

kT

m, (6.44)

so

vt ≈√

3 cs. (6.45)

Since cs ∼ 300m s−1, vt ∼ 500m s−1. As a check, let’s do the cal-

culation directly. To convert between different energy units, multiply

(6.43) by unity:

vt =

√

3kT

m× c

c=

√

3kT

28mpc2c. (6.46)

Now the energies in the square root have convenient values:

mpc2 ≈ 109 eV,

kTroom ≈ 1

40eV.

(6.47)

The electron–Volts cancel and

vt ∼(

3

40 × 28 × 109

)1/2

c ∼ 1

6·10−5c = 500m s−1. (6.48)

2006-01-22 12:49:46 [rev 251865795177]

6. Thermal properties 95

Area σ

ℓ

Figure 6.6. Mean free path. A molecule

has scattering area σ, which is πd2.

Once it sweeps out a volume contain-

ing one molecule, then it is likely tocollide. The distance λ that makes the

cylinder contain one molecule is then

the mean free path. Its volume is σλ, so

the number of molecules it contains is

nσλ. Hence nσλ ∼ 1.

To compute λ see Figure 6.6. It is determined by the molecular

diameter d and by the number density n:

nσλ ∼ 1, (6.49)

here σ is the scattering cross section. You can also recreate this re-

lation by ensuring that the dimensions match: n has dimensions of

1/volume, σ is an area, and λ is a length, so the left side has no di-

mensions, and neither does the right side. To compute n you need σ,

which is πd2. Why is it πd2 rather than the more intuitive area of a

circle πr2? Because two molecules collide if their centres come within

a distance d = 2r. So you can consider the moving molecule as having

a radius R = 2r and all the other molecules being points. Then the

relevant cross-sectional area is πR2 or πd2.

To compute λ for air, we need its molecular diameter. Air is N2.

Guessing that λ ∼ 3 A or d ∼ 3 A is always reasonable. Then

λ ∼ 1

nσ∼ 2.4 ·10−2 m3

6 ·1023︸ ︷︷ ︸

n

× 1

3︸︷︷︸

π

× 9 ·10−20 m2

︸ ︷︷ ︸

d2

∼ 10−7 m

(6.50)

and

κ ∼ 1

3× 500m s−1

︸ ︷︷ ︸

vt

× 10−7 m︸ ︷︷ ︸

λ

∼ 1.5 ·10−5 m2 s−1. (6.51)

In decently sized units (cgs units),

κ ∼ 0.15 cm2 s−1. (6.52)

That value is suspiciously similar to the viscosity of air (4.45). The

connection is not accidental. Thermal diffusivity is the diffusion con-

stant for heat. Kinematic viscosity is the diffusion constant for mo-

mentum. In a gas, molecular motion carries heat and momentum, so

their respective diffusion constants are closely related. Their ratio is

dimensionless. It is such an important number that it is given a name,

the Prandtl number:

Pr ≡ ν

κ, (6.53)

It is close to unity in gases and even in many solids and liquids (for

water, it is 6).

Example 6.2 Sniffing perfume

You open a bottle of perfume at one end of a room. How longbefore people at the other end can smell it? If you make enough ap-proximations, you can apply the preceding results. Pretend that per-fume molecules are merely labelled air molecules (imagine that they

2006-01-22 12:49:46 [rev 251865795177]

6. Thermal properties 96

Substance cp/k

I2 4.4

Cl2 4.1

O2 3.5

N2 3.5

Ni 3.1

Au 3.1

Zn 3.1Fe 3.0

Xe 2.5

He 2.5

Diamond

0 ◦C 0.6

223 ◦C 1.6

823 ◦C 2.6

Table 6.4. Specific heats at constant

pressure. All data are for room temper-

ature (unless otherwise noted) and at-mospheric pressure. The specific heats

are in dimensionless form – in units of

k per atom or molecule – because we al-

ready know that the specific heat must

contain a factor of k. Source: Smithso-

nian Physical Tables [17, p. 155].

are green, although that makes no physical sense). How long beforea green air molecule wanders across the room? The time is t ∼ l2/κ,where l is the length of the room, say 5m. Then

t ∼(5m)2

1.5 ·10−5 m2 s∼ 106 s. (6.54)

A day is 105 s, so the green wandered takes 10 days to reach the othernoses. This estimate is far too high! The method ignores a few minoreffects. For example, noses are extremely sensitive, and can detect evena handful of molecules (for certain molecules), but that effect turns outto reduce t by only a small factor, perhaps 5. The method also ignoresthe size and mass of perfume molecules. They are much heavier andlarger than air molecules, which slows their diffusion. Perhaps thateffect cancels the factor from nose sensitivity. Even if it does not fullycancel the factor, the resulting t is still far too long. But do not spendtoo much time correcting the small mistakes. Otherwise you fall intoa trap [1]:

In other words – and this is the rock solid principle on whichthe whole of the Corporation’s Galaxy-wide success is founded– their fundamental design flaws are completely hidden by theirsuperficial design flaws.

By agonizing over the superficial flaws, you will the fundamental flaw,which is that the perfume odors are transported mostly by convection.The room is full of tiny air currents, due to temperature differencesbetween the floor and ceiling, due to people walking, due to doorsopening, etc. These currents, whjich are minature winds, move perfumemolecules much more efficiently than diffusion does. The calculation ofthe diffusion time is correct, and is not a total waste because it showsyou that odors must travel using another mechanism.

6.5 Specific heat of gases

The thermal conductivity K requires the thermal diffusivity, the sub-

ject of the previous section, and the specific heat, the subject of this

section. How much energy does it take to heat water to bath tem-

perature? How many days of solar heating can the oceans store? The

answers to these questions depend on the energy that the substance

stores per unit temperature change: the specific heat.

Before thinking about the physics of specific heats, make the usual

dimensional estimate. The dimensions of specific heat are

[specific heat] =energy

temperature × amount of substance. (6.55)

The amount can be whatever size is convenient: one mole, one gram,

or one molecule. The molecule is a natural size: It involves the fewest

arbitrary parameters. A mole, or a gram, for example, depends on

human-chosen sizes. You know one quantity with units of energy per

temperature: the Boltzmann constant. So a first estimate is

specific heat

molecule∼ k,

specific heat

mole∼ kNA = R.

(6.56)

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6. Thermal properties 97

Table 6.4 lists the specific heat of various substances.

For nitrogen (or air),

cp =7

2k, (6.57)

A useful way to remember k as 0.8× 104 eV K−1 and convert it to SI

units when you need to:

k ≈ 0.8 eV

104 K× 1.6 ·10−19 J

eV= 1.3 ·10−23 J K−1. (6.58)

So

cp = 4.5 ·10−23 J K−1. (6.59)

Now you have the ingredients to compute K for air (which is mostly

nitrogen). The number density is

n =6 ·1023

1mol× 1mol

24 ℓ× 1 ℓ

10−3 m3= 2.5 ·1025 m−3. (6.60)

With κ from (6.51) and cp from (6.59), the thermal conductivity is

K = ncpκ ∼ 2.5 ·1025 m−3 × 4.5 ·10−23 J K−1 × 1.5 ·10−5 m2 s−1

∼ 1.7 ·10−2 W m−1 K−1.(6.61)

Now imagine air at the same temperature but different density (and

pressure). For concreteness, imagine decreasing the density by a factor

of 4 and consider the effect on each factor in the product K = ncpκ.

The number density decreases by a factor of 4. The cp is the specific

heat per molecule, so it does not change. What about κ? As a result

of n decreasing, the mean free path, which is l ∼ 1/nσ, increases

by a factor of 4. Since the diffusivity is κ ∼ vtl, and vt remains the

same (T is constant), the diffusivity increases by a factor of 4. The

product ncpκ therefore does not change, so thermal conductivity is

independent of density over a wide range of densities – and this result

holds not just for air, but for any ideal gas. Once the mean free path

gets comparable to the size of the container, however, this analysis

breaks down. One system with a huge container is the atmosphere.

Figure 6.7 shows how little K varies despite ρ varying over many

orders of magnitude. A similar result holds for the dynamic viscosity

η = ρν. For ideal gases, ν ∼ κ. So ν inherits the ρ−1 dependence from

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6. Thermal properties 98

•

•

•

•

•

•

•

•

•

10−5 10−4 0.001 0.01 0.1 1

0.018

0.02

0.022

0.024

ρ (kg m−3)

K (W m−1 K−1)

Figure 6.7. Thermal conductivity vs

density. The data are from the ‘U.S.

Standard Atmosphere (1976)’ at various

heights up to 80 km. The density variesover five orders of magnitude, whereas

the thermal conductivity varies by 30

percent. Source: CRC [41, pp. 14-20ff].

κ (which got it from the mean free path), and η = νρ is a constant.

6.6 Diffusivity of liquids and solids

Many of these ideas for gases apply to liquids and solids. Most impor-

tant, the formulas for flux and thermal conductivity are the same. So

to find the thermal conductivity K, it is a matter of finding cp and

κ. In a liquid or a solid, heat is not transported by molecular motion.

Heat is molecular motion but, unlike in a gas, it is not transported

by molecular motion. In a solid, for example, the molecules hardly

leave their sites. They vibrate but do not wander much. In a liquid,

molecules wander but only slowly: The tight packing means that the

free path is very short. Yet everyday experience suggests that liquids

and solids can be excellent conductors of heat. The reason is that heat

is transported by sound waves or phonons rather than by molecu-

lar motion. One molecule vibrates, shaking the next molecule, which

shakes the next molecule, and so on.

Using this idea we can compute a thermal diffusivity. Heat is the

vibration of atoms. In a solid, the atoms are confined in a lattice, and

the vibrations can be represented as combinations of many sound

waves. More precisely, the waves are phonons, which are sound waves

that can have polarization (just as light waves have polarization).

Heat diffusion is the diffusion of phonons.

Phonons act like particles: They travel through lattice, bounce

off impurities, and bounce off other phonons. The same random-walk

analysis for gases works for solids and liquids, perhaps with a different

velocity or mean free path. The phonon mean free path λ measures

how far a phonon travels before bouncing (or scattering), and then

heading off in a random direction. In a solid without too many defects,

at room temperature, a phonon travels a few lattice spacings before

scattering. Let’s say

λ = fa, (6.62)

where a is the lattice spacing and f is a dimensionless number that

2006-01-22 12:49:46 [rev 251865795177]

6. Thermal properties 99

says how many lattice spacings the phonons survives. Now for the

speed. In a gas the relevant speed for computing κ was the speed at

which the molecules move; this speed is roughly the sound speed. In

a liquid or solid, the phonon speed is the relevant speed, and it is also

the sound speed, but it has no relation to the thermal speed. To see

why, recall the sound speed (5.61):

cs ∼√

ǫc

m, (6.63)

where ǫc is the cohesive energy and m is the atomic or molecular

mass. A typical thermal speed is vt ∼√

kT/m, so the dimensionless

ratio cs/vt is

cs

vt

∼√

ǫc

m

/√

kT

m=

√ǫc

kT. (6.64)

At room temperature, kT ∼ 1/40 eV. For a typical solid, ǫc ∼ 10 eV.

So the ratio is√

400 or 20. Compare that value to its counterpart in

an ideal gas:cs

vt

∼ 1√3

(ideal gas). (6.65)

With the mean free path (6.62), the thermal diffusivity is

κ =1

3csfa. (6.66)

For most solids or liquids, cs ∼ 3 km s−1 (see Table 5.5 on p. 80) and

a ∼ 3 A. Then

κ ∼ 1

3× 3 ·103 m s−1

︸ ︷︷ ︸

cs

× f × 3 ·10−10 m︸ ︷︷ ︸

λ

∼ f

3× 10−6 m2 s−1. (6.67)

For a typical solid, take f = 3. Then

K ∼ 10−6 m2 s−1 = 10−2 cm2 s−1. (6.68)

For our favorite substance, water, cs ∼ 1.7 km s−1 and a ∼ 3 A, so

κ ∼ f × 1.7 ·10−7 m2 s−1 (water). (6.69)

In cgs units,

κ ∼ f × 1.7 ·10−3 cm2 s−1. (6.70)

For water, the value is exact if you take f = 1, an easy result to

remember. So phonons travel only one lattice spacing before being

scattered. This short distance is because water has no lattice, so the

molecules are highly disordered. The many irregularities scatter the

phonons. Let’s look at two examples of thermal diffusion.

Example 6.3 Cooking a turkey

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6. Thermal properties 100

t ∝ m0.58

•

•

•

•

•

•

8 9 10 12 14 16 18 20 25

3

3.5

4

4.5

5

m (lb)

t (hrFigure 6.8. Turkey cooking times (log–

log plot). The times are typical recom-

mended times. The nontraditional units

for mass are appropriate, since the ma-

jor application is for an American hol-

iday. These data are taken from [49]

(for an unstuffed turkey). The red lineis the best power-law fit to the data. Its

exponent of 0.58 is not far from the pre-

diction of 2/3.

During Thanksgiving, an American holiday in which large turkeysare cooked, many people give thanks once all the turkey leftovers arefinished. How long does it take to cook a typical turkey? Consider aspherical turkey with, say, r ∼ 10 cm. A turkey is roughly water, sothat radius would have m ∼ 4 kg, reasonable for feeding a family of10. The time for heat to diffuse from the edge, where the air and panare hot, to the center, is roughly t ∼ r2/κ. Using the cgs value (6.70)for κ:

t ∼(10 cm)2

1.7 ·10−3 cm2 s−1∼ 6 ·104 s, (6.71)

or roughly 15 hours! This estimate is too large, perhaps by a factor of4. Still, start cooking in the morning if you want dinner before mid-night. Even though the missing dimensionless constant is significant,the derivation suggests an extension. Imagine turkeys of varying size ormass: How does the cooking time vary? This extension calls for a scal-ing analysis. The size is r ∝ m1/3 so the cooking time is t ∝ r2

∝ m2/3.Figure 6.8 shows that this analysis is reasonably accurate.

The estimate of the cooking time ignored an important parame-ter: the oven temperature. So the result implicitly assumes that theturkey will cook in the time t no matter what the oven temperatureis. Diffusion equalizes the temperature at the center of the turkey tothe oven temperature. Equalization is necessary but not sufficient. Theinside must cook (the proteins must denature) after reaching the oventemperature. So that is why the oven must be hot: A cold oven wouldnot cook the meat, even after the meat reached oven temperature.What oven temperature is hot enough? Protein physics is complicatedso this temperature is difficult to estimate by theoretical calculationswhile sitting an armchair, so use experimental data. From experiencepan frying a thin filet of fish, you can guess that a thin piece of meatnext to a hot skillet (perhaps at 200 ◦C) cooks in a few minutes. If theskillet is only at 50 ◦C, the meat will never cook (50 ◦C is not muchhotter than body temperature). So with the oven at 200 ◦C, the centerof the turkey turkey will cook in a few minutes after it reaches oventemperature. So as long as the oven is hot enough, most of the cook-ing time is spent attaining this temperature, which is the assumptionbehind the estimate of r2/κ for the cooking time.

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6. Thermal properties 101

Example 6.4 Cooling the moon

How long does the moon take to cool? Its size is r ∼ 2 · 106 m,as you estimated in (2.5). For the thermal diffusivity of rock, use thetypical value (6.67). What f is reasonable? Rock is highly disordered,so f (the number of lattice spacings that the phonons travel) is notlarge – at most perhaps 3. Then κ ∼ 10−6 m2 s−1. So

t ∼(2 ·106 m)2

10−6 m2 s−1∼ 4 ·1018 s ∼ 1011 yr. (6.72)

[Another useful number: 1 yr ∼ 3 · 107 sec.] The universe has beenaround for only 1010 yr; so why is the moon already cold? This prob-lem is similar to the paradox in the perfume example (Example 6.2):How does a person at the other end of the room smell the perfumeif molecules takes 10 days to diffuse there? In both cases the answeris little winds (convection). Rock seems like a solid, but it flows ona billion-year timescale, and this flow transports heat to the surfacemuch faster than molecular motions could.

This example shows a merit of order-of-magnitude physics: effi-ciency. You could – with lots of computer time – solve the diffusionequation for a mostly-spherical body like the moon. You might showthat the accurate cooling time is, say, 2.74 · 1011 yr. What would thatresult gain you? The more accurate time is still far longer than theage of the universe, so the same paradox remains. It has the sameconclusion: convection. Order-of-magnitude analysis allows you to de-termine quickly which approaches are worth pursuing, so you can thenspend more time refining productive approaches and less time chasingunproductive ones.

6.7 Specific heat of liquids and solids

The remaining piece in the thermal conductivity is the specific heat,

either cp or cv. Liquids and solids hardly change volume, even if the

volume is not held constant, so the two specific heats are almost

equal. If each molecule sits in a three-dimensional harmonic potential

produced by the rest of the lattice, each molecule has three potential-

energy degrees of freedom. Combining them with the three transla-

tional degrees of freedom produces six degrees of freedom. Each degree

of freedom contributes kT/2 to the internal energy so the energy per

molecule is 3kT . Since cp ∼ u/T :

cp ∼ 3k. (6.73)

The specific heat per mole is

Cp ∼ 3R ∼ 25J

mol K. (6.74)

which is a useful number to remember. This value is the lattice heat

due to phonons transporting energy. The prediction is accurate for

most of the solids and liquids (Ni, Au, Zn, and Fe) in Table 6.4.

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6. Thermal properties 102

K T

Substance W m−1 K−1 ◦C

Diamond 1000 0

H2O

water 0.6071 25

snow 0.16 0

ice 2.3 −5Glass

borosilicate crown 1.1 25

pyrex 1.1 25

Wood

oak 0.16 20

plywood 0.11 20

pine, ⊥ 0.11 60

pine, ‖ 0.26 60

Wool 0.04 30

Rock

granite 2.2 20limestone 1 20

basalt 2 20

Cork 0.06 0

Asphalt 0.06 20

Sand, dry 0.33 20

Concrete 0.8 0

Ethanol 0.169 25

Quartz

‖ to c-axis 11 25

⊥ to c-axis 6.5 25

Paper 0.06 25Asbestos 0.09 0

Table 6.5. Thermal conductivities for

common liquids and solids. The table

uses the SI unit for K, not only because

it is a standard unit but also because it

is a typical size for K as predicted in

(6.75). Tabulate values relative to a rea-

sonable estimate, so that typical values

come out near unity. Source: Kaye and

Laby [30, §2.3.7] and the CRC [41, 6-

190, 12-224ff].

6.8 Thermal conductivity of solids and liquids

Now the pieces are assembled to estimate a typical K:

K ∼(

1

3 A

)3

︸ ︷︷ ︸

n

× 3 × 1.3 ·10−23 JK−1

︸ ︷︷ ︸

cp

× 10−6 m2 s−1

︸ ︷︷ ︸

κ

∼ 1W m−1 K−1.

(6.75)

Table 6.5 lists useful thermal conductivities.

6.9 Thermal conductivity: Metals

A curious fact in Table 6.4 is that nickel, gold, and zinc have cp slightly

greater than 3k! From where did the extra specific heat come? These

substances are all metals, so the estimate in (6.74) probably neglects

an important feature of metals. Metals are different from other solids

because their electrons are free to roam and to absorb thermal energy.

Therefore the total specific heat should include the specific heat of

the electrons. Their contribution is small, enough to raise cp by 0.1k

above the prediction of 3k.

But this small contribution does not explain why metals feel colder

to the touch than, say, concrete does. If phonons are the only mech-

anism for conducting heat, then metals and concrete would not have

such a great disparity in conductivity. That they do is a consequence

of the large contribution to the thermal conductivity from free elec-

trons in a metal. The typical phonon contribution (6.75) is K ∼1W m−1 K−1. Rather than computing the electron contribution from

scratch, scale it relative to the phonon contribution. Because

conductivity = diffusivity × specific heat, (6.76)

the problem of finding the ratio of conductivities becomes the problem

of finding the ratio of diffusivities and of specific heats. First, the ratio

of electronic to phonon diffusivities. Diffusivity is

velocity × mean free path, (6.77)

so split the estimate of the diffusivity ratio into estimates of the ve-

locity ratio and of the mean-free-path ratio.

Phonons move at speeds similar to the speed of sound – a few

kilometers per second. Electrons move at the Fermi velocity vF.

Electrons are confined by electrostatics (a metal is like a giant hy-

drogen atom). In both systems, confinement gives electrons a velocity

and a kinetic energy. Let ne be the number density of free electrons

in the metal. Then ∆p ∼ ~/∆x ∼ ~n1/3e , where ∆p is the momen-

tum uncertainty produced by confinement, and ∆x is the position

uncertainty. The velocity ∆p/me is the Fermi velocity:

vF ∼ ~n1/3e

me

. (6.78)

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6. Thermal properties 103

pF

Figure 6.9. Fermi sphere. It is a sphere

in momentum space with radius pF,which is the Fermi momentum mevF. If

an electron absorbs a packet of energy,

it gets a new momentum and moves to a

new point in the sphere. However, most

points are not accessible. The lower-

energy states in the sphere are filled

and the Pauli principle will not allow an

electrons to join the others in that state.

So it has to go to an empty state. But

if kT ≪ EF, only electrons in the tiny

ring beyond the dashed ring have enoughenergy to move to an empty state. So

most electrons are immune from thermal

fluctuations.

A typical metal has one or two free electrons per atom. Pretend it

is one electron per atom. Then n1/3e = 1/a, where a = 3 A is the

interatomic spacing. To evaluate vF, multiply by unity:

vF ∼ c~cn

1/3e

mec2

∼ 3 ·105 kms−1 ×

~c︷ ︸︸ ︷

2000 eV A×

n1/3e

︷ ︸︸ ︷

0.3 A−1

5 ·105 eV︸ ︷︷ ︸

mec2

∼ 500 km s−1.

(6.79)

This velocity is much faster than a typical sound wave, which may

have cs ∼ 2 or 3 km s−1. So the velocity ratio is ∼ 200 (electrons win).

Not only are electrons faster than phonons, but for reasons related to

the Fermi surface, they travel farther before scattering than phonons

do. In copper, λe ∼ 100a, where a ∼ 3 A is a typical lattice spacing.

The phonon mean free path is λp ∼ 10 A. So the mean-free-path ratio

is ∼ 30. The diffusivity ratio is therefore 200 × 30 ∼ 104 in favor of

the electrons.

Now for the specific-heat ratio. The phonons win this race be-

cause only a small fraction of the electrons carry thermal energy. To

understand why, revisit the Fermi-velocity argument. A more accu-

rate picture is that the metal is a three-dimensional potential well

(the electrons are confined to the metal). There are many energy lev-

els in the well, labeled by the momentum of the electrons that occupy

it. The free electrons fill the lowest levels first. The Pauli principle

allows only two electrons to share an energy level. The electrons in

the highest levels have velocity vF or momentum mevF. As vectors,

the momenta of the highest-energy electrons are uniformly distributed

over the surface of a sphere in momentum space; the sphere has radius

mevF and is called the Fermi sphere (Figure 6.9). To carry thermal

energy, an electron has to be able to absorb and deposit energy; the

typical package size is kT . Consider an electron that wants to absorb

a thermal-energy package. When it does so, it will change its energy

– it will move outward in the Fermi sphere. But can it? If it is in

most of the sphere, it cannot, because the sphere is packed solid – all

interior levels are filled. Only if the electron is near the surface of the

sphere can it absorb the package. How near the surface must it be?

It must have energy within kT of the Fermi energy EF (the Fermi en-

ergy is ∼ mev2F). The fraction of electrons within this energy range is

f ∼ kT/EF. Typically, kT ∼ 0.025 eV and EF ∼ few eV, so f ∼ 10−2.

This fraction is also the specific-heat ratio. To see why, consider the

case f = 1 (when every electron can carry thermal energy). Each atom

contributes 3k to the specific heat, and each electron contributes one-

half of that amount (3k/2) because there is no spring potential for

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6. Thermal properties 104

Substance K(102 W m−1 K−1

)

Ag 4.27

Al 2.37

Cr 0.937

Cu 4.01

Fe 0.802

Ni 0.907

W 1.74Hg 0.0834

Table 6.6. Thermal conductivities for

common metals at room temperature.

Metals should have thermal conductiv-

ities of roughly 102 W m−1 K−1, so we

measure the actual values in that unit.

We are being gentle with our neural

hardware, giving it the kind of num-

bers that it handles with the least dif-

ficulty: numbers close to 1. The good

electrical conductors (copper, silver, and

gold) have high thermal conductivities.In both cases, electrons do the trans-

porting, either of charge (for electrical

conductivity) or of heat (for thermal

conductivity), and the electron mean free

path is long. Mercury, the only liquid

in the table (in red), has a low thermal

conductivity, because the lack of a lattice

shortens the electron mean free path to

only one or two lattice spacings.

the electrons (they contribute only their three translational degrees of

freedom). The number of free electrons is typically twice the number

of atoms, so the total electron and phonon contributions to specific

heat are roughly equal. When f ∼ 10−2, the contributions have ratio

10−2.

Now assemble all the pieces. The conductivity ratio is

Kmetal

Kdielectric

∼ 104

︸︷︷︸

κ

× 10−2

︸ ︷︷ ︸

ncp

= 102. (6.80)

No wonder metals feel much colder than glass or wood! The typical

dielectric thermal conductivity (6.75) is 1W m−1 K−1. So

Kmetal ∼ 102 W m−1 K−1. (6.81)

Table 6.6 contains data on the thermal conductivities of common

metals. The estimate (6.81) is more accurate than it has a right to

be, given the number of approximations that made in deriving it.

6.10 What you have learned

Approximate first, check later: Many analyses are simpler if a di-

mensionless parameter is small. So assume that it is small, as we

did when computing the thermal expansion in Figure 6.4, arrive

quickly at a result, and then use it to check whether the assump-

tion was justified.

The microscopic basis of thermal diffusivity and viscosity: Parti-

cles (or phonons) move in steps whose size is equal to the mean

free path, λ. The particles’ velocity v determines the time to take

one step and the diffusion constant, which is κ ∼ vλ.

Viscosity: For gases, kinematic viscosity ν and thermal diffusiv-

ity κ are approximately the same. In other words, the ratio ν/κ,

known as the Prandtl number, is roughly 1.

Formula hygeine: Convert formulas to dimensionless form, as for

the calculation of thermal expansion, so that equations are simpler

and reasoning is easier.

Metals: In metals, electrons transport most of the heat; phonons

store most of the heat.

Typical values: Typical thermal conductivities are

K ∼{

0.02 for air,1 for non-metals,100 for metals.

× W m−1 K−1 (6.82)

6.11 Exercises

◮ 6.28 Losing heat

At what rate do you lose heat when standing outside in a T-shirt on

a winter day?

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6. Thermal properties 105

◮ 6.29 Wooden spoons

Why are wooden spoons useful for cooking?

◮ 6.30 Eggs

How long should it take to hardboil an egg?

◮ 6.31 Sweating

How much water do you sweat away while bicycling at your full speed

for 10 km?

◮ 6.32 Earth cooling

How long should the earth take to cool? Interpret your estimate.

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